Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
To calculate the amount of heat transferred when an amount of reactant is decomposed, we must look at the balanced reaction and its corresponding heat of reaction. In this case, we can see that 252.8 kJ of heat is transferred per 2 moles of CH3OH used. When 22 g of CH3OH is used, 86.9 kJ is absorbed.
Explanation:
As it is known that there are two types of properties. These are extensive and intensive.
Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.
Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.
On the basis of these properties water and ethanol are distinguished as follows.
- Density of water is 997 kg/
whereas density of ethanol is 789 kg/. Both these liquids can be separated by intensive properties. - Melting point of water is zero degree celsius whereas melting point of ethanol is -114.1 degree celsius.
- Specific heat capacity of water is 4.184
whereas specific heat capacity of ethanol is 2.46 . - Mass of the given liquids cannot be differentiated because they will keep on changing depending on the quantity required. As mass is an extensive property, therefore, it is difficult to differentiate between the two liquids.
Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.
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